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Old 09-07-2006, 08:10 PM   #61
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infinity plus anything = infinity. it's not a higher or lower infinity. its the same infinity. because either way its going on forever no matter how much you take away or add or divide by... unless you divide infinity by infinity. that's 1. or divide infinity by 0, can't be done. or subtract infinity from infinity, that's 0.
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Old 09-07-2006, 08:11 PM   #62
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if the limit comes out to be .999999 then its just 1.
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Old 09-07-2006, 09:08 PM   #63
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Quote:
Originally Posted by DreamRuled
very good...that's why this is a theoretical question and answer. lyou can't actually have an infinite number of zeros lined up, so you can't have a one at the end. but in theory, it is possible to say that after the infinte number of zeros, there will be a one, much like saying that after the infinite number of reals following 1, there is a 2. slightly different circumstance, but similar enough.
But on the 1 and 2...


You are giving "restrictions" to your set of infinity...

You are saying "between" 1 and 2 ONLY....So the infinity between 1 and 2 yes you can...

But this infinity, (and all others where you don't add "restrictions") numbers don't go on the end...

There are numbers after the infinity between 1 and 2...But there aren't after infinity...
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Old 09-07-2006, 09:09 PM   #64
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Quote:
Originally Posted by geeWiz15
infinity plus anything = infinity. it's not a higher or lower infinity. its the same infinity. because either way its going on forever no matter how much you take away or add or divide by... unless you divide infinity by infinity. that's 1. or divide infinity by 0, can't be done. or subtract infinity from infinity, that's 0.
Yeah thats what I was saying...

So Geewiz...

Agree or disaggee that 0.9999 repeat = 1?


If no...

Explain the 1/3 situation


And what is 1 - 0.9 repeat
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Old 09-07-2006, 09:10 PM   #65
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Quote:
Originally Posted by Hawker
if the limit comes out to be .999999 then its just 1.
There you go

But you can go look at the case why it doesn't equal too...

Because, well, its just not there yet...
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Old 09-07-2006, 09:10 PM   #66
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Quote:
Originally Posted by XxNeXuSxX
This is debated all the time, and really can't be answered.


If you have a full sandwich, eat one half of it, eat another half, eat another half.

Keep eating it in half, will the sandwich ever be finished?


Mathematically no, engineers would usually say yes.



What is this, the gourmand's version of Xeno's paradox?

The "problem" is actually similar to those encountered in asymptotic calculus, but in practice it's clearly a simple one to solve. Look to the philosophy of Democritus for an answer.

Last edited by Loki : 09-07-2006 at 09:12 PM.
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Old 09-07-2006, 09:12 PM   #67
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Of course, in math it's just like an asymptotic line, get's closer, but will never reach to (0), simple math solution.



It's just the different ways it's interpreted that I find interesting.
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Old 09-07-2006, 09:14 PM   #68
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What kind of sandwich?
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Old 09-07-2006, 09:17 PM   #69
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Quote:
Agree or disaggee that 0.9999 repeat = 1?
no

Quote:
Explain the 1/3 situation
.333 repeating forever is just the numerical representation. it will never actually be equal to the true value of 1/3... it will keep on going forever and never actually approach the true value of 1/3. .333 is approaching the asymptote of 1/3 but never actually touches it just like .999 repeating does 1.

Quote:
And what is 1 - 0.9 repeat
a number that gets closer and closer to the asymptote of 0 as can be done as the decimals go on without actually touching it.
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Old 09-07-2006, 09:18 PM   #70
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Well the smallest piece of matter as we take it is the atom, so you would split it up to the last atom, then you would split the last atom of the sandwich and then blow up.
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Old 09-07-2006, 09:29 PM   #71
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Default OT: Interesting Calculus Phenomenon

usually when given a particular complex function you need to integrate and you cant, you need to rely on integration by parts. eventually you will get a product minus an integral with an easily antidifferentiable function.


however, sometimes there is a phenomenon when if you continue to integrate by parts, the original integral appears again as you integrate. durin these circumstances, you can create a reduction formula in which you can easily calculate the integral of any family of functions which share a common attribute:


for instance:

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Old 09-07-2006, 09:56 PM   #72
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The site where you got that from is stupid. The integral can be evaluated in five lines using a simple substitution.
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Old 09-07-2006, 10:11 PM   #73
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okay smart guy


integrate (secx)^3 in five lines using a simple substitution. :rollingeyes:
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Old 09-07-2006, 10:37 PM   #74
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Quote:
.333 repeating forever is just the numerical representation. it will never actually be equal to the true value of 1/3... it will keep on going forever and never actually approach the true value of 1/3. .333 is approaching the asymptote of 1/3 but never actually touches it just like .999 repeating does 1.


So then what is the true value of 1/3 that 0.33 repeating keeps approaching?
There is no answer to that from you, because
0.3 repeating IS the true value of 1/3

Once again, do you know how to divide numbers like that on paper? Like 3/5, do you know how to do that on paper? It will give you 0.6...Now, do 1/3 on paper, and suprise, 0.33 will start coming up, forever...

Quote:
a number that gets closer and closer to the asymptote of 0 as can be done as the decimals go on without actually touching it.

That makes no sense at all...

Your acting like 0 is the only number it can't equal...So can it equal 0.2?
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Old 09-07-2006, 11:11 PM   #75
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The answer is 1 but here is my reason.

I've been thining about this for awhile. Check it out.

1/9 is .1 repeating
2/9 is .2 repeating
3/9 is .3 repeating and also 1/3
.
.
etc
.
.
so, 9/9 is .9 repeating... and also from fractions we know that to be 1.

Good post.
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